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Motion Graphs

Part of the Mechanics section of Coddy's Physics journey. Lesson 29 of 97.

On a v-t graph the slope is the acceleration and the area under the line is the displacement.

A velocity-time graph rising from 0 to 12 metres per second in 4 seconds, flat for 4 seconds, then falling to 0 at 10 seconds; the area under it is split into a triangle of 24 metres, a rectangle of 48 metres and a triangle of 12 metres

Slope of the v-t line = acceleration; area under it = displacement.

Under a straight sloped line the area is the average velocity times the time:

Δx=v0+v2 t\Delta x = \frac{v_0 + v}{2}\, t

Area below the time axis counts as negative displacement.

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Challenge

A cyclist's v-t graph is a straight line from 2 m/s2\ \mathrm{m/s} at t=0t = 0 to vv at t=6 st = 6\ \mathrm{s}. In that time she covers 48 m48\ \mathrm{m}. Find her final velocity vv.

Try it yourself

Physics diagram: plot

Solve for vv

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